Chapter 4.4, Problem 6CFU

To determine

To list: The possible rational roots of equation $2x^3+3x^2-8x+3=0$ and then determine the rational roots.

Answer to Problem 6CFU

Possible rational roots: $\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

The actual rational roots: $1,-3\text{ and }\frac{1}{2}$

Explanation of Solution

Given information: $2x^3+3x^2-8x+3=0$

Calculation:

Given function

  $2x^3+3x^2-8x+3$

Rational Root Theorem says in order to find all possible rational roots you must divide the factors of both p and q, where p = the constant term and q = the leading coefficient.

The function must be in standard form (highest degree to lowest degree).

Leading coefficient $\left(q\right)=2$

Constant term $\left(p\right)=3$

Factors of $q=\pm 1,\pm 2$

Factors of $p=\pm 1,\pm 3$

After dividing each factors these are all the possible rational roots of the function.

Factors of $\begin{array}{l}\frac{p}{q}=\frac{\pm 1,\pm 3}{\pm 1,\pm 2}\\ =\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}\end{array}$

At this point, plotting the function on a graph can help find zeros since it will help approximate which ones actually make sense. I tried x = 1 and using synthetic division ended up with 0, so 1 is one of the rational zeros.

Synthetic division leaves you with the coefficients of a new function, which you can use to find the roots as well.

  

  $2x^2+5x-3$

New function from synthetic division

Factored

Solve for x

  $\begin{array}{l}2x^2+5x-3\\ \left(2x-1\right)\left(x+3\right)\\ x=\frac{1}{2}\text{ and }x=-3\end{array}$

Possible rational roots: $\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

The actual rational roots: $1,-3\text{ and }\frac{1}{2}$